Question

Question 3: A natural gas has the following molar analysis: CH4, 80. 62%; C2H6, 5, 41%; C3H8, 1. 87%; C4H10, 1. 60%; N2, 10. 50%. The gas is burned with dry air, giving products having a molar analysis on a dry basis: CO2 7. 8% CO, 0. 2%; 02, 7% N2, 85%. (a) Determine the air - fuel ratio on a molar basis. (b) Assuming ideal gas behavior for the fuel mixture, determine the arrount of products in kmol that would be formed from 100 m3 of fuel mixture at 300 K and 1 bar. (c) Determine the percent of theoretical air. For help the hint is given as: The solution can be conducted on the basis of an assumed amount of fuel mixture or on the basis of an assurmed amount of dry products. Let us illustrate the first procedure, basing the solution on 1 kmol of fuel mixture. The chemical equation then takes the form (0. 8062CH4 + 0. 0541C2H6 + 0. 0187C3H8 + 0. 0160C4H10 + 0. 1050N2 + a (O2 + 3. 76N2) - b (0. 078C02 + 0. 002C0 + 0. 0702 + 0. 85N2) + H2O a = theoretical number of moles of air required b = the number of moles of CO2, CO, N2 formed c = The number of moles of H2O formed

Answer 1

Data Provided in the problem statement

CH4 = 80. 62 %

C2H6 = 5.41%

C3H8 = 1. 87%

C4H10 = 1. 60%

N2 = 10. 50%

Burning of fuel reaction takes the form

[ (0. 8062 CH4 + 0. 0541 C2H6 + 0. 0187 C3H8 + 0. 0160 C4H10 + 0. 1050 N2 ] + a ( O2 + 3. 76 N2) ==>

b (0. 078 C02 + 0. 002 C0 + 0. 07 O2 + 0. 85 N2) + c H2O

Here

a =Number of moles of air required

b = moles of products formed

c = moles of water formed

Now taking carbon balance on above reaction , we get

[ (0. 8062 CH4 + 0. 0541 C2H6 + 0. 0187 C3H8 + 0. 0160 C4H10 + 0. 1050 N2 ] + 2.8914 ( O2 + 3. 76 N2) ==>

12.931 (0. 078 C02 + 0. 002 C0 + 0. 07 O2 + 0. 85 N2) + 1.9295 H2O

Now Air- Fuel ration is given as

AF = ( moles of air ) / ( moles of fuel) = ( 2.8914) ( 1 + 3.76 ) / 1 = 13.763 kmol air / kmol fuel

B)

Now kmol fuel in 100 m3 of fuel at 300 K and 1 bar  is calculated by ideal gas equation as

C)

For complete combustion of fuel reaction can be written as,

[ (0. 8062 CH4 + 0. 0541 C2H6 + 0. 0187 C3H8 + 0. 0160 C4H10 + 0. 1050 N2 ] + ( O2 + 3. 76 N2) ==>

C02 + N2  + H2O

Balancing this reaction we get,

[ (0. 8062 CH4 + 0. 0541 C2H6 + 0. 0187 C3H8 + 0. 0160 C4H10 + 0. 1050 N2 ] + 2.0 ( O2 + 3. 76 N2) ==>

1.0345 C02 + 7.625 N2  + 1.93 H2O

Now theoretical fuel-air ratio is given as

AF = ( moles of air ) / ( moles of fuel) = ( 2) ( 1 + 3.76 ) / 1 = 9.52 kmol air / kmol fuel

Percent theoretical air is given as

Percent theoretical air = ( ( Actual AF ratio ) / ( theoretical AF ratio ) ) * 100

= ( ( 13.763 kmol air / kmol fuel ) / ( 9.52 kmol air / kmol fuel ) ) * 100

Percent theoretical air = 144.56 %